Obtain the differential equation for a LC circuit.

Explanation

L-C circuit is shown in figure. In their circuit capacitor (C ) and inductor ( L ) are connected. Let at time t=0, capacitor is charged and charge on it is `q_(m)`.

The moment the circuit is comleted the charge on the capacitor starts decreasing giving rise to current in the circuit.
Let q and I be the chrage and current in the circuit at time t.
Since `( dI)/( dt) ` is positive the induced emf in L will have polarity as shown in figure,
Means `V_(a) gt V_(b)`.
As q decreases I increases,
` :. I = - (dq)/(dt)`
Induced emf in inductor at any isntant,
`V = epsilon = - L (dI)/(dt)`
and p.d. across capacitor `= ( q)/( C )`
According to Kirchhoff.s loop rule,
`- L (dI)/(dt) + ( q)/( C ) = 0`
but `I = - ( dq)/( dt)`
`:. (dI)/(dt) = - ( d^(2)q)/( dt^(2))`
`:. + L ( d^(2) q)/( dt^(2)) + ( q)/( C ) = 0 `
`:. (d^(2) q)/( dt^(2)) + ( q)/( LC ) = 0 `
`:. (d^(2)q)/( dt^(2)) + omega_(0)^(2) q= 0 ` `[ :. (1)/(LC) = omega_(0)^(2) ]`
which is second order differential equation of linear charged.
This equation has the form of general formula, for a simple harmonic oscillator.The charge therefore, oscillates with a natural frequency.
`omega_(0) = (1)/( sqrt(LC))`
`:. 2 pi f_(0) = (1)/( sqrt(LC))`
`:.` Its frequency is `f_(0) = (1)/(2pi ) sqrt((1)/( LC))`